[sequin]

FWIW, I just gave Deepseek the same prompt and it solved it too (much faster than the 41m of ChatGPT). I then gave both proofs to Opus and it confirmed their equivalence.

The answer is yes. Assume, for the sake of contradiction, that there exists an \(\epsilon > 0\) such that for every \(k\), there exists a choice of congruence classes \(a_1^{(k)}, \dots, a_k^{(k)}\) for which the set of integers not covered by the first \(k\) congruences has density at least \(\epsilon\).

For each \(k\), let \(F_k\) be the set of all infinite sequences of residues \((a_i)_{i=1}^\infty\) such that the uncovered set from the first \(k\) congruences has density at least \(\epsilon\). Each \(F_k\) is nonempty (by assumption) and closed in the product topology (since it depends only on the first \(k\) coordinates). Moreover, \(F_{k+1} \subseteq F_k\) because adding a congruence can only reduce the uncovered set. By the compactness of the product of finite sets, \(\bigcap_{k \ge 1} F_k\) is nonempty.

Choose an infinite sequence \((a_i) \in \bigcap_{k \ge 1} F_k\). For this sequence, let \(U_k\) be the set of integers not covered by the first \(k\) congruences, and let \(d_k\) be the density of \(U_k\). Then \(d_k \ge \epsilon\) for all \(k\). Since \(U_{k+1} \subseteq U_k\), the sets \(U_k\) are decreasing and periodic, and their intersection \(U = \bigcap_{k \ge 1} U_k\) has density \(d = \lim_{k \to \infty} d_k \ge \epsilon\). However, by hypothesis, for any choice of residues, the uncovered set has density \(0\), a contradiction.

Therefore, for every \(\epsilon > 0\), there exists a \(k\) such that for every choice of congruence classes \(a_i\), the density of integers not covered by the first \(k\) congruences is less than \(\epsilon\).

\boxed{\text{Yes}}

[CGamesPlay]

> I then gave both proofs to Opus and it confirmed their equivalence.

You could have just rubber-stamped it yourself, for all the mathematical rigor it holds. The devil is in the details, and the smallest problem unravels the whole proof.

[yosefk]

How dare you question the rigor of the venerable LLM peer review process! These are some of the most esteemed LLMs we are talking about here.

[Davidzheng]

"Since \(U_{k+1} \subseteq U_k\), the sets \(U_k\) are decreasing and periodic, and their intersection \(U = \bigcap_{k \ge 1} U_k\) has density \(d = \lim_{k \to \infty} d_k \ge \epsilon\)."

Is this enough? Let $U_k$ be the set of integers such that their remainder mod 6^n is greater or equal to 2^n for all 1<n<k. Density of each $U_k$ is more than 1/2 I think but not the intersection (empty) right?

[Paracompact]

Indeed. Your sets are decreasing periodic of density always greater than the product from k=1 to infinity of (1-(1/3)^k), which is about 0.56, yet their intersection is null.

This would all be a fairly trivial exercise in diagonalization if such a lemma as implied by Deepseek existed.

(Edit: The bounding I suggested may not be precise at each level, but it is asymptotically the limit of the sequence of densities, so up to some epsilon it demonstrates the desired counterexample.)

[Klover]

Here's kimi-k2-thinking with the reasoning block included: https://www.kimi.com/share/19bcfe2e-d9a2-81fe-8000-00002163c...

[nsoonhui]

I am not familiar with the field, but any chance that the deepseek is just memorizing the existing solution? Or different.

https://news.ycombinator.com/item?id=46664976

[utopiah]

Sure but if so wouldn't ChatGPT 5.2 Pro also "just memorizing the existing solution?"?

[logicchains]

Opus isn't a good choice for anything math-related; it's worse at math than the latest ChatGPT and Gemini Pro.

[amluto]

I find it interesting that, as someone utterly unfamiliar with ergodic theory, Dini’s theorem, etc, I find Deepseek’s proof somewhat comprehensible, whereas I do not find GPT-5.2’s proof comprehensible at all. I suspect that I’d need to delve into the terminology in the GPT proof if I tried to verify Deepseek’s, so maybe GPT’s is being more straightforward about the underlying theory it relies on?